A car with mass m-1000 kg completes a turn of radius r 540 m at a constant speed of v =25 m/s. As the car goes around the turn, the tires begin to slip. Assume that the turn is on a level road, i.e. the road is not banked at an angle. Randomized Variables r 540 m v=25 m/s What is the numeric value of the coefficient of static friction, us, between the road and tires?

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aristocles aristocles · Answer 1 · 2020-04-03T07:02:54+02:00

Answer:

friction coefficient on the rough road is 0.117

Explanation:

As we know that car is taking turn on flat rough road

So here we can say that required centripetal force for circular motion of car is due to frictional force

So we have

[tex]F_c = \frac{mv^2}{R}[/tex]

now we have

[tex]\mu mg = \frac{mv^2}{R}[/tex]

so we have

[tex]\mu = \frac{v^2}{Rg}[/tex]

[tex]\mu = \frac{25^2}{540(9.81)}[/tex]

[tex]\mu = 0.117[/tex]

abidemiokin abidemiokin · Answer 2 · 2021-09-27T14:13:10+02:00

The coefficient of static friction, between the road and tires is 0.118

The centripetal force acting on the car is expressed as:

[tex]F_c=\frac{mv^2}{r}[/tex]

Since the centripetal force acting on the car bring about friction between the tyre and the road, hence;

[tex]\mu R=\frac{mv^2}{r} \\\mu mg = \frac{mv^2}{r} \\\mu g = \frac{v^2}{r}[/tex]

Given the following

r = 540m

v = 25m/s

g = 9.8m/s²

Substitute the given values into the formula;

[tex]9.8\mu = \frac{25^2}{540} \\9.8 \mu =\frac{625}{540}\\9.8 \mu = 1.1574\\\mu = \frac{1.1574}{9.8} \\\mu = 0.118[/tex]

Hence the coefficient of static friction, between the road and tires is 0.118

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